INA592: using INA592 for making a HOWLAND PUMP

This is the last simulation that I can make with Pspice. Now the problem is that changing the resistors  the result doesn't change. and it always gives the maximum current of 65mA.

Hi Farzaneh,

Enclosed is an example with Iout = 33mA using OPA592. The Iout is set from Vdelta/Iout, where Vdelta = 2.5V and Iout = 33mA --> RL = 2.5V/33mA = 75.76Ω 

OPA592 33mA Howland Ipump 09172023.TSC

Farzaneh Faridi said:

I need to make a part of my design to be a HOWLAND PUMP for driving up to 100mA.

In order to generate +/-100mA current source, the op amp needs to be capable to source/sink the required current (unless current booster is used). Our OPA593 is well suited for the application, see the simulation below. 

The OPA593 can handle up to +/-42.5Vdc or up to 85Vdc in power rails, and is capable to source/sink up to 250mA. The gain resistor ratios need to be matched better than +/-1% in order to improve CMRR. 

OPA593 Howland PWM +-100 mA Ipump 09202023.TSC

If you have other questions, please let me know. 

Best,

Raymond

Dear Raymond,

Thanks for your answer. As you mentioned, I was thinking to use a current booster like a BJT for increasing the current. The one that ypou sent me with opa 593 is really great. What I need to know exactly is the gain of the circuit. In your design it's 5 right? 

Thanks

Hi Faraneh, 

The OPA593 difference amplifier has gains of 5V/V. The part also has current limit capability and may work well for the application. 

If you'd like to have other constant current option, please let me know. 

Best,

Raymond

I want to make a calculation for calculating the output current. I changed the Rl to 25 and R1 to 50 the current was 220A. then I changed Rl to 100 and R1 to 200 the current was 58.51A. 

On the other hand I just run the simulation and the results are different from what it is inserted as an image. Am I doing sth wrong?

Best

Hi Farzaneh, 

There are output swing limitation in OPA593 and you have to check with the datasheet. 

OPA593 Howland PWM +-200 mA Ipump 09202023.TSC

By changing the RL from 50 to 25ohm, the current will double, so it Vout at the load at the output current source (IR drop). I reduced the R1 from 100 to 60 ohm and it seems to work for the load. One of the option is to increase the supply rails from +/-18Vdc to higher voltage rails. If the Iout does not require to be bidirectional, then you may use single supply rail and increase its voltage ratings.  

When op amp's Iout is increased, the IR voltage drop across the mosfet drivers also increase, and this will limit the output voltage swing as shown in the plots below. 

If you have other questions, please let me know.

Best,

Raymond

Dear Ratmond,

I was unable to make the equation of the gain and the resistances. I'm trying to make a calculator for further changes, based on the input voltage and the sense and load resistances and the gain.

I know that the gain of the circuit is 5, but i need to know to calculate it. Using the simulation I fpund out the input voltage does not make any effect as we expect it from a Howland current pump. Then I see that changing the power supply of the operational changes , maintaining the resistores values, the output current changes.

Thanks

Farzaneh

Thanks a lot